1. Prep Book Chapter 5 – Definition pf the Derivative
Looking to calculate the slope of a tangent line to a curve at a particular point.
Use formulas to construct functions that we then calculate a limit for.
The limit we calculate IS the value of the SLOPE OF THE TANGENT LINE.
Two formulas based on y/x

2. = Definition 1 for slope of tan. line.
Slope Definition 1 y x
f(x) - f(a)
m = =
x - a f
f(x) - f(a)
x - a
lim x-->a
Q(x, f(x))
[f(x) - f(a)]
= Definition 1 for slope of tan. line.
P(a, f(a))
(x - a) x a

3. ex: Find the equation of the tangent line to the curve y = x2 @ point P = (1,1)
f(a)
a = 1, f(a) = 1
f(x) - f(a)
x - a
lim x-->a x2
- 1
(x + 1)(x - 1)
(x - 1)
= lim x-->1
= lim x-->1
x - 1
= lim x-->1
(x + 1) = 2 = m
Point-Slope Form: [y - y1 = m(x - x1)]
= [y - 1 = 2(x - 1)]
y - 1 = 2x - 2
y = 2x - 1
= equation of the tangent line

4. = Definition 2 for slope of tan. line.
Slope Definition 2
m = y x =
f(a+h) - f(a) h f
f(a+h) - f(a) h
lim h-->0
Q(a+h), f(a+h))
[f(a+h) - f(a)]
= Definition 2 for slope of tan. line.
P(a, f(a)) h
(a + h) a

5. ex: Find the equation of the tangent line to the hyperbola y = 3
ex: Find the equation of the tangent line to the hyperbola y =
@ point P = (3, 1) x
f(a) = 1
a = 3
a+h = 3+h 3
(3+h)
= lim
h-->0
3 - (3+h)
(3+h) h - 1
f(a+h) - f(a) h
lim h-->0
= lim h-->0 h
= lim
h-->0 -h
(3+h) h
= lim
h-->0 -h
(3+h) x h 1
= lim
h-->0 -1
(3+h) = -1 3 1

6. 2.6 cont’d - Velocities
Ex: Suppose a ball is dropped from a position 450m high. The equation of its motion is (s) = f(t) = 4.9t2
What is the velocity of the ball after 5 seconds?
How fast is the ball traveling when it hits the ground? s
(5, 122.5)
122.5 t 5

10. Calc 2.8 - The Derivative as a Function
f(a+h) - f(a) h
f’(a) = lim
h-->0
f(x+h) - f(x) h
f’(x) = lim
h-->0
so…
If we continuously vary x, f’(x) will also vary….
so if we look at the original function f, then f’(x) is the slope of function f at a particular point x…
so if we match up a particular point x on function f with the value of the slope of f at that point x, we have point (x, f’(x))…
which means we can PLOT A NEW CURVE using x and using f’(x) as y-values….
which means we can now refer to f’(x) as a separate function…(that can be graphed just like f(x) can)…. 10

11. ex:
f(x)
f’(x) 11

12. ex: f(x) = x3 - x f’(x) = 3x2 - 1 1
ex: f(x) = x f’(x) =
2 x 12

13. 2.8 continued…
Differentiation - The process of calculating a derivative. For standard notation y = f(x), derivative notation may be:
f ’(x)
y ’
dy/dx (Leibniz notation)
d/dx f(x)
Definition of “Differentiable” - A function is differentiable at a # ‘a’ if f ’(a) exists. It is differentiable on an open interval if it is differentiable at every number in the interval… (a,b) , [a,b], (a,), etc…

14. Continuity vs. Differentiability
If f is differentiable at ‘a’, then f is continuous at ‘a’.
If f is continuous at ‘a’, f is NOT necessarily differentiable at ‘a’.
Non-Differentiable Characteristics:
Kinks
Discontinuities
Vertical Tangents

15. f (x) is a function and may have its own derivative (f )  = f 
The Second Derivative
f (x) is a function and may have its own derivative
(f )  = f 
Leibniz notation: d dx dy =
d2y
dx2
ex:
For f(x) = x3 - x and f (x) = 3x2 - 1, find and interpret f (x)…
f(x+h) - f(x) h
f (x) = lim
h-->0

16. 3(x + h)2 - 1 - (3x2 - 1) h lim = f (x) = 3x2 - 1
3(x2 + 2xh + h2) (3x2 - 1) h
lim
h-->0 =
3x2 + 6xh + 3h x2 + 1 h
lim
h-->0 =
6xh + 3h2 h
lim
h-->0 =
lim 6x + 3h
h-->0 = 6x
= f (x)
Interpretation: Any derivative is a rate of change - therefore the “derivative of a derivative” is “a rate of change of a rate of change”… the second derivative is the rate at which the first derivative is changing.

17. f(x)
f(x)
f(x)
f(x)
f(x)
f(x) 17

18. Derivative Chain of Position
Velocity
Acceleration
s(t)
s’(t) = v(t)
s’’(t) = v’(t) = a(t) ds dt dv dt
d2s
dt2 or
Leibniz Notation:
Acceleration = instantaneous rate of change of velocity.